Class 9 Science Exploration Chapter 7 Solutions (NCERT 2026–27) – Work, Energy, and Simple Machines

These Class 9 Science Exploration Chapter 7 solutions cover Work, Energy, and Simple Machines from the new NCF-2023 textbook (2026–27).

Class: 9 Subject: Science Book: Exploration Chapter: 7 Exercise: Revise, Reflect, Refine (15 Qs) Session: 2026–27
On this page

Class 9 Science Exploration Chapter 7 Solutions – Overview

Chapter 7 of Exploration, Work, Energy, and Simple Machines, explains the scientific meaning of work (force × displacement), the two main forms of mechanical energy — kinetic energy and potential energy — the work–energy theorem, the law of conservation of energy, the meaning of power, and the working of simple machines such as the lever. These Class 9 Science Exploration Chapter 7 solutions answer every textbook question step by step.

Key Concepts & Definitions

Work: done when a force moves an object through a displacement in the direction of the force; W = Fs (zero if there is no displacement).

Energy: the capacity to do work; SI unit joule (J).

Kinetic energy: energy of a moving body, KE = ½mv2. Potential energy: energy due to position/height, PE = mgh.

Conservation of energy: energy can change form but the total energy stays the same.

Power: the rate of doing work, P = W/t (SI unit watt, W).

Simple machine (lever): balances when load × load arm = effort × effort arm (principle of moments).

Work, Energy & Power Formulas

Work: W = F × s (joule).

Kinetic energy: KE = ½mv2.

Potential energy: PE = mgh.

Work–energy theorem: work done by the net force = change in kinetic energy.

Power: P = W/t (watt); and v2 = u2 + 2as for motion under gravity.

“Think It Over” — Answers

Two children slide down slides of different shapes but the same height. Will they reach the bottom with the same speed?

ANSWEROn a smooth (frictionless) slide the speed at the bottom depends only on the height dropped: v = √(2gh). Since both slides have the same height, both children reach the bottom with the same speed, whatever the shape.

Which slide gives the greatest speed at the bottom?

ANSWERAll slides of the same height give the same final speed. A steeper slide only makes the child reach the bottom in less time (greater average acceleration), not at a greater final speed.

Class 9 Science Exploration Chapter 7 Solutions — Revise, Reflect, Refine

1. State whether True or False. (i) Work is said to be done when a force is applied, even if the object does not move. (ii) Lifting a bucket vertically upward results in positive work done on the bucket. (iii) The SI unit for both work and energy is joule (J). (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.

ANSWER (i) False — work needs displacement; no movement means no work. (ii) True — the force and displacement are both upward, so the work is positive. (iii) True — both are measured in joules. (iv) False — a stretched but motionless rubber band has elastic potential energy, not kinetic energy. (v) True — energy can be transformed from one form to another.

2. Fill in the blanks. (i) Work done = ______ × ______ (in the direction of force). (ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force. (iii) The expression for kinetic energy of a body of mass m and velocity v is ______. (iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ______. (v) Power is defined as the ______ at which work is done.

ANSWER (i) force × displacement. (ii) 1 (one) newton. (iii) ½mv2. (iv) mgh. (v) rate.

3. When a ball thrown upwards reaches its highest point, tick which statement(s) are correct. (i) The force acting on the ball is zero. (ii) The acceleration of the ball is zero. (iii) Its kinetic energy is zero. (iv) Its potential energy is maximum.

ANSWER Correct: (iii) and (iv). At the highest point the velocity is zero, so the kinetic energy is zero and the potential energy is maximum. (i) and(ii) are wrong: gravity still acts, so the force is mg (downward) and the acceleration is g.

4. For each situation, identify the energy transformation: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi) explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric bulb, (ix) a solar panel.

ANSWER (i) chemical energy (fuel) → kinetic + potential energy. (ii) elastic potential energy → kinetic (mechanical) energy. (iii) light (solar) energy → chemical energy. (iv) potential energy → kinetic energy. (v) chemical energy → heat + light energy. (vi) chemical energy → heat + light + sound + kinetic energy. (vii) sound energy → electrical energy. (viii) electrical energy → light + heat energy. (ix) light (solar) energy → electrical energy.

5. A student (mass m = 50 kg) is lifted straight up in a lift to the top of a building of height h = 72.5 m, then later climbs the stairs to the same top (g = 10 m s-2). (i) Find the gain in potential energy if the student is lifted straight up. (ii) Find the gain in potential energy when the student climbs the stairs. (iii) What do you conclude about the dependence of potential energy on the path taken?

ANSWER (i) PE = mgh = 50 × 10 × 72.5 = 36250 J. (ii) The height gained is the same, so PE = mgh = 36250 J again. (iii) The gain in potential energy depends only on the vertical height, not on the path taken.

6. A crane lifts a mass m to the 10th floor in a certain time, then raises the same mass to the 20th floor in double the time. How much more energy and power are required? (All floors are of equal height.)

ANSWER Energy ∝ height. The 20th floor is twice the height of the 10th floor, so the energy needed is twice as much (E2 = 2E1). Power = energy/time. Here energy doubles but the time also doubles, so P2 = 2E1/2t = E1/t = P1 — the power required is the same.

7. Which factors determine the energy required to raise a flag to the top of a flagpole using a pulley? Does raising it slowly or quickly change the work done? If the speed is doubled, how does the power requirement change?

ANSWER The work (energy) depends only on the weight of the flag (mg) and the height of the pole (h): W = mgh. Raising the flag slowly or quickly does not change the work done — it is the same either way. But power = work/time. If the speed is doubled, the time is halved, so the power required doubles.

8. A man (60 kg) rides a scooter (100 kg) and accelerates it to a velocity v. The next day his son (40 kg) joins as a passenger and the scooter reaches the same speed in the same time. What is the ratio of the fuel used on the two days? (All energy goes to the scooter; ignore air resistance and friction.)

ANSWER Energy needed = kinetic energy at speed v = ½mv2, so it is proportional to the total mass. Day 1 mass = 60 + 100 = 160 kg; Day 2 mass = 160 + 40 = 200 kg. Ratio of fuel = 160 : 200 = 4 : 5.

9. On a seesaw, a child sits on one side and an adult (weight twice the child’s) on the other, yet the seesaw is balanced. Draw a figure showing the distances of the child and the adult from the fulcrum.

ANSWER For balance (principle of moments): weight of child × its distance = weight of adult × its distance. Since the adult’s weight is twice the child’s, the adult must sit at half the distance of the child from the fulcrum — the child sits at distance 2d and the adult at distance d. Figure (in words): a horizontal beam on a triangular fulcrum; child at distance 2d on the left, adult (twice the weight) at distance d on the right, so 1×2d = 2×d balances.

10. A ball of mass 2 kg is thrown up with a velocity of 20 m s-1. (i) Identify the sign of the work done by gravity during the upward and downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance? (g = 10 m s-2)

ANSWER (i) Upward motion: gravity acts opposite to the displacement, so its work is negative. Downward motion: gravity acts along the displacement, so its work is positive. (ii) Initial KE = ½mv2 = ½(2)(202) = 400 J. PE gained at 19.4 m = mgh = 2 × 10 × 19.4 = 388 J. Energy lost = 400 − 388 = 12 J, so the work done by air resistance = −12 J (it removed 12 J of energy).

11. A 10.0 kg block moves on a near-frictionless floor. A variable force acts in its direction of motion from 0 m to 4 m as in Fig. 7.37. If the block had 180 J of kinetic energy at 0 m, find its speed (i) at 0 m, (ii) at 4 m. Does it have negative acceleration anywhere?

ANSWER (i) KE at 0 m = 180 J ⇒ ½(10)v2 = 180 ⇒ v2 = 36 ⇒ v = 6 m s-1. Work done = area under the force–displacement graph = ½(1)(50) + (2)(50) + ½(1)(50) = 25 + 100 + 25 = 150 J. (ii) KE at 4 m = 180 + 150 = 330 J ⇒ ½(10)v2 = 330 ⇒ v2 = 66 ⇒ v = 8.12 m s-1. The applied force stays in the direction of motion (always positive) throughout 0–4 m, so the acceleration is never negative — the block keeps speeding up.

12. The gravitational attraction on the Moon’s surface is about one-sixth of that on Earth. An astronaut can throw a ball up to a height of 8 m on Earth. How high will the same throw go on the Moon?

ANSWER For the same upward velocity, maximum height h = v2/(2g), so h is inversely proportional to g. On the Moon g is one-sixth of Earth’s, so the height becomes 6 times: hMoon = 6 × 8 = 48 m.

13. A 1000 kg car moving at constant speed brakes to a stop. Its speed–time graph from when the driver spots the obstruction is shown in Fig. 7.38. (i) Describe how the car moves between positions A and B. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes between B and C. (iv) What does the kinetic energy of the car transform into?

ANSWER (i) From A to B (0–1 s) the speed is constant at 35 m s-1 — the car moves with uniform velocity (the driver’s reaction time, before braking). (ii) KE at A = ½mv2 = ½(1000)(352) = 612500 J (6.125 × 105 J). (iii) From B to C the brakes bring the car to rest, so they do work equal to the whole kinetic energy: −612500 J (magnitude 6.125 × 105 J). (iv) The kinetic energy is transformed mainly into heat (and a little sound) in the brakes.

14. The potential energy–displacement graph of a 0.5 kg ball on a frictionless track is shown in Fig. 7.39. At O the velocity is 0 m s-1 and the potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

ANSWER Total mechanical energy is conserved: E = KE + PE = 0 + 30 = 30 J. So at any point KE = 30 − PE and v = √(2 KE / m) = √(4 KE). At P (PE = 20 J): KE = 30 − 20 = 10 J ⇒ v = √(4 × 10) = √40 = 6.32 m s-1. At Q (PE = 30 J): KE = 30 − 30 = 0 ⇒ v = 0 m s-1 (a turning point). At R (PE = 40 J): this needs 40 J but only 30 J is available, so KE would be negative — the ball cannot reach R (it does not have enough energy).

15. A coconut of mass 1.5 kg falls from a 10 m tall tree onto wet sand and stops by making a depression. (g = 10 m s-2) (i) Calculate the velocity of the coconut just before it hits the sand. (ii) If the average resistive force of the sand is 3000 N and all the coconut’s energy makes the depression, calculate its depth.

ANSWER (i) v = √(2gh) = √(2 × 10 × 10) = √200 = 14.14 m s-1. (ii) Energy at impact = mgh = 1.5 × 10 × 10 = 150 J. This equals the work done by the sand: F × d = 150. Depth d = 150 / 3000 = 0.05 m = 5 cm.

Common Mistakes to Avoid

Watch out for these

  • Thinking work is done whenever a force acts — there must be a displacement in the direction of the force.
  • Confusing kinetic energy (½mv2) with momentum (mv), or potential energy (mgh) with weight (mg).
  • Forgetting that potential energy depends only on height, not on the path.
  • Mixing up energy and power — doing the same work faster needs more power, not more energy.
  • Dropping the negative sign for work done by gravity (going up) or by a resistive/braking force.
  • Not using conservation of energy for graph problems (KE + PE = constant on a frictionless track).

Extra Practice Questions

Very Short Answer Type Questions

Q1. State the SI unit of power.

ANSWERThe watt (W); 1 W = 1 J s-1.

Q2. What kind of energy is stored in a stretched bow?

ANSWERElastic potential energy.

Q3. When is the work done by a force zero even though the force acts?

ANSWERWhen there is no displacement, or the displacement is perpendicular to the force.

Short Answer Type Questions

Q1. A body of mass 5 kg moves at 4 m s-1. Find its kinetic energy.

ANSWERKE = ½mv2 = ½(5)(42) = ½(5)(16) = 40 J.

Q2. A pump raises 200 kg of water to a height of 10 m in 20 s. Find its power. (g = 10 m s-2)

ANSWERWork = mgh = 200 × 10 × 10 = 20000 J. Power = W/t = 20000/20 = 1000 W (1 kW).

Long Answer Type Question

Q1. Using a freely falling body, show that the total mechanical energy is conserved.

ANSWER At the top (height h, at rest): KE = 0, PE = mgh, total = mgh. After falling a distance x: speed v2 = 2gx, so KE = ½m(2gx) = mgx, and PE = mg(h − x); total = mgx + mg(h − x) = mgh. Just before hitting the ground (fallen h): KE = mgh, PE = 0, total = mgh. The total energy stays mgh throughout, so mechanical energy is conserved.

MCQs & Assertion–Reason

1. The SI unit of work is the:

(a) watt    (b) newton    (c) joule    (d) pascal

2. The kinetic energy of a body of mass m moving with speed v is:

(a) mv    (b) ½mv2    (c) mgh    (d) mv2

3. The work done is zero when the angle between force and displacement is:

(a) 0°    (b) 45°    (c) 90°    (d) 180°

4. The SI unit of power is the:

(a) joule    (b) watt    (c) newton    (d) kelvin

5. Water stored in a dam has mainly:

(a) kinetic energy    (b) potential energy    (c) heat energy    (d) sound energy

6. If the speed of a body is doubled, its kinetic energy becomes:

(a) double    (b) half    (c) four times    (d) unchanged

7. 1 kilowatt-hour equals:

(a) 1000 J    (b) 3600 J    (c) 3.6 × 106 J    (d) 60 J

8. A body raised to height h has potential energy:

(a) mgh    (b) ½mv2    (c) mg    (d) mh

9. On a frictionless track, as a ball rolls down, its potential energy:

(a) increases    (b) converts to kinetic energy    (c) becomes heat    (d) stays constant

10. A lever is balanced when:

(a) load = effort    (b) load × load arm = effort × effort arm    (c) load arm = effort arm    (d) load > effort

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(b), 6-(c), 7-(c), 8-(a), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: No work is done in carrying a bag while walking on a level road.

Reason: The force (upward) is perpendicular to the horizontal displacement.

A-R 2. Assertion: If the speed of a body doubles, its kinetic energy becomes four times.

Reason: Kinetic energy is proportional to the square of the speed.

A-R 3. Assertion: The gain in potential energy is the same whether you climb stairs or take a lift to the same floor.

Reason: Potential energy depends only on the vertical height raised.

A-R 4. Assertion: Doing the same work in a shorter time needs more energy.

Reason: Power is the rate of doing work.

A-R 5. Assertion: A ball at the highest point of its path has zero kinetic energy.

Reason: Its velocity is zero at the highest point.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • Work = force × displacement (in the direction of the force); unit joule.
  • Kinetic energy = ½mv2; potential energy = mgh.
  • Work–energy theorem: net work = change in kinetic energy.
  • Energy is conserved — it only changes form.
  • Power = work/time; unit watt; 1 kWh = 3.6 × 106 J.
  • A lever balances when load × load arm = effort × effort arm.

Real-life Applications

These ideas explain everyday life: hydroelectric dams convert the potential energy of stored water into electricity, roller coasters trade height for speed, our food gives us energy to do work, levers and pulleys make lifting easier, and power ratings (watts) tell us how fast appliances use energy — which is what the electricity bill (in kWh) charges for.

How to score full marks in this chapter

Write the correct formula (W = Fs, KE = ½mv2, PE = mgh, P = W/t) with units, and watch the sign of work. For graph problems use area = work (force–displacement) and KE + PE = constant on frictionless tracks. Distinguish energy (total) from power (rate).

Frequently Asked Questions

What is Class 9 Science Exploration Chapter 7 about?

Work, energy and simple machines — work done by a force, kinetic and potential energy, the work–energy theorem, conservation of energy, power, and levers.

What are the main formulas in this chapter?

W = F × s, KE = ½mv2, PE = mgh, and P = W/t.

Does potential energy depend on the path taken?

No. It depends only on the vertical height raised, so a lift and the stairs to the same floor give the same gain in potential energy.

Are these Class 9 Science Exploration Chapter 7 solutions free?

Yes. All solutions are free and follow the official NCERT Exploration textbook for 2026–27.

← Chapter 6 All Exploration Chapters Chapter 8 →
Scroll to Top